what does this mean: is (a molecule) sigma or pi in symmetry? the molecule in question is c triple bond c. ....it has 2 pi bonds and 1 sigma bond....which is symmetrical?

I don't know what that means. The acetylene molecule is a linear molecule which makes it symmetrical in two dimensions but not in three. I suggest you Google sigma bonds and pi bonds and look at the pictures. If we knew what kind of symmetry the question was talking about we might do a better job of answering.

Read the intro statement at the top of the page then click on some of the links below that for more information.

https://www.google.com/search?q=are+pi+bonds+symmetrical&ie=utf-8&oe=utf-8

thanks haha it's a strange question to ask

When determining the symmetry of a molecule, we need to understand the types of chemical bonds involved. In this case, the molecule in question is C≡C, which consists of a triple bond between two carbon atoms.

The sigma (σ) bond is formed by the overlap of two atomic orbitals head-on, along the line connecting the two nuclei. In C≡C, the sigma bond is formed by the overlap of two sp hybrid orbitals from each carbon atom. Since there is only one sigma bond in C≡C, we refer to it as a sigma bond.

On the other hand, the pi (π) bond is formed by the sideways overlap of p orbitals, which are perpendicular to the molecular axis. In C≡C, there are two pi bonds, which result from the overlap of two sets of parallel p orbitals—one from each carbon atom. Each pi bond consists of two parallel electron clouds above and below the sigma bond.

Now, back to the question of symmetry. The sigma bond remains symmetrical because it is formed by the head-on overlap of orbitals along the molecular axis. However, the pi bonds are not symmetrical. Each pi bond involves the sideways overlap of p orbitals, resulting in two parallel electron clouds above and below the molecular axis. These electron clouds are not symmetrical around the molecular axis.

Thus, in the molecule C≡C, the sigma bond is symmetrical, but the pi bonds are not.